i started reading about templates and i got confused on the below. -
template<class T>
-
T max(T t1, T t2)
-
{
-
if (t1 > t2)
-
return t1;
-
return t2;
-
}
-
int main(){
std::cout<<max< int>120,14.55);//=>as per my understanding here if i explicitly mention only one data type,other would be deducted from the argument type(14.55) by the compiler.That means T max(T t1, T t2) instantiates T max(int t1,double t2)in this case.But when i compiled it i got the below warning
warning: passing double for argument 2 to T max(T, T) [with T = int].Then why this warning came is my question since i have already instantiated for T max(int t1,double t2).
also since we are using only T,if i mention max<int>,then T max(T t1, T t2) should be like this int max(int t1, int t2) na??
return 0;
}
Please clear my doubts.else i cant proceed further
1 2065
Your template only uses T as a parameter. That means you can't use int and double.
Since the template only returns T you can't return int or double.
A template is only a pattern. When you specialize the template, the compiler makes a copy of it and replaces the T with the type you specify. This creates the function that will be called.
You use explicit specialization then certain types require different logic: - template <class T>
-
bool IsZero(T arg)
-
{
-
if (arg) return true;
-
return false;
-
}
-
//explicit
-
//do not use the template when T is double
-
bool IsZero(double arg)
-
{
-
if (fabs(arg) < 0.0001) return false;
-
return true;
-
}
Ths tells the compiler to not use the template if T is double.
Since bool IsZero(double arg) is an actual function, it will not be in a header file but will be in an implementation file somewhere in the build. Therefore, you need to tip off the compiler that the function exists. You do it this way: - template <class T>
-
bool IsZero(T arg)
-
{
-
if (arg) return true;
-
return false;
-
}
-
//explicit specialization
-
//do not use the template when T is double
-
template<> bool IsZero(double arg);
-
}
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